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09 Jun 2011, 08:42

The pattern depends where the 2 blue chips will be placed in the row (the others will inevitably be red). To place the first blue chip, we have 5 possibilities, for the second one 4, so this would give 5x4=20 possibilities. But there is no difference in pattern regarding the two blue chips (they are identical), therefore, we should divide by 2, so there are 5x4/2=10 different possible patterns. The correct answer is A.
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13 Jun 2012, 04:52

Where the formula is giving the answer to be 10 number of ways to arrange the chip, I am getting 11 ways when arranging them manually. Those 11 ways of arrangements are as follows.BBRRRBRBBRBRBRBBRRBRBRRRBRBBRRRBRBRRBRRBRRBBRRRBRBRRRBB

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Where the formula is giving the answer to be 10 number of ways to arrange the chip, I am getting 11 ways when arranging them manually. Those 11 ways of arrangements are as follows.BBRRRBRBBRBRBRBBRRBRBRRRBRBBRRRBRBRRBRRBRRBBRRRBRBRRRBB

This problem can be approached in two ways: 1. Total number of arrangements/(arrangements of R's & S's) = 5!/3!2! = 10Which is basically removing the arrangements from permutations to get the combinations, or,2. Total number of ways in which 3 R's can be arranged at 5 places, rest of the places would be filled by B,or 5C3 = 10

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13 Jun 2012, 06:28

Whenever you have a question like this with two identical sets, think of it as 5! where you have a total of 5x4x3x2x1= 120 different sets of rearrangements. But you also have 3 identical red chips and 2 identical blue chips so you MUST account for these as well.

5!/3!2!= 5x4x3x2x1/ 3x2x1x2x1 which equals out to 10 different arrangements being possible.

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Whenever you have a question like this with two identical sets, think of it as 5! where you have a total of 5x4x3x2x1= 120 different sets of rearrangements. But you also have 3 identical red chips and 2 identical blue chips so you MUST account for these as well.

5!/3!2!= 5x4x3x2x1/ 3x2x1x2x1 which equals out to 10 different arrangements being possible.

Would this be the most efficient approach on the GMAT?

THEORY FOR SUCH KIND OF PERMUTATION QUESTIONS:

Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:

\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).

For example number of permutation of the letters of the word "gmatclub" is \(8!\) as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).

BACK TO THE ORIGINAL QUESTION:There are 3 red chips and 2 blue chips. If they form a certain color pattern when arranged in a row, for example RBRRB, how many color patterns are possible?A. 10B. 12C. 24D. 60E. 100

According to the above the # of permutations of 5 letters BBRRR of which 2 B's and 3 R's are identical is \(\frac{5!}{2!*3!}=10\).

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14 Jun 2013, 00:39

somebody asked above y would we divide by 3! and 2!.

Its because the 3 reds and 2 blues are identical on their own.if the anagram would have been r(1)r(2)r(3)b(1)b(2) (the numbers in paranthesis is to denote the subscripts for each ) the anagram r(1)r(2)r(3)b(1)b(2) would be diff compared to r(2)r(1)r(3)b(1)b(2) .But here the case is identical .So we have to remove the redundant one's.

There fore arranging them in 5! and later dividing by 3! and 2! to remove the redundant cases.